[Bug c/53369] Integral promotion with ~ operator - Different behavior for signed and unsigned char.

["pinskia at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=53369

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|                            |INVALID

--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> 2012-05-16 00:11:53 UTC ---
    signed char a = 1 << 7;
    unsigned char b = 1 << 7;

    printf("%hd  %hu\n", ~a, ~b);


let's see ~a is really ~(int)a.  Likewise ~b is really ~(int)b.
#define printf __builtin_printf
int main(void)
{    
    signed char a = 1 << 7;
    unsigned char b = 1 << 7;

    printf("%x  %x\n", ~a, ~b);
}

--- CUT ----
7f  ffffff7f


This is the correct behavior.  As (int)(unsigned char)a does a zero extend as
the unsigned char fits directly in an int.

Think of this way.  The value from unsigned char and signed char are unchanged
when promoted into int.  So with the first one you get -128 and the second case
you get 128.

and then you take the ~ and you get 127 and (~(1<<7)) .